Answer:
Center Of Mass lies at 5L/9 from O as shown.
Explanation:
The Center of mass of any object is given by
[tex]X_{com}=\frac{\int xdm}{\int dm}\\\\Y_{com}=\frac{\int ydm}{\int dm}[/tex]
where (x,y) is location of point with mass dm
Now since rod is a linear object we have it's y co-ordinates as zero
Thus
[tex]COM_{rod}=\frac{\int xdm}{\int dm}\\\\[/tex]
The mass of an element 'dx' at a distance 'x' from O can be written as
[tex]dm=\lambda(x) dx\\\\\lambda(x) ={\lambda _0}+\frac{2\lambda _{0}-\lambda _{0}}{L}x\\\\\lambda (x)=\lambda _0+\frac{\lambda _0}{L}x[/tex]
Thus we have
[tex]dm=\lambda(x) dx\\\\\lambda(x) ={\lambda _0}+\frac{2\lambda _{0}-\lambda _{0}}{L}x\\\\\lambda (x)=\lambda _0+\frac{\lambda _0}{L}x\\\\COM_{rod}=\frac{\int x\lambda(x)dx}{\int \lambda(x)dx}\\\\=\frac{\int (\lambda_{0}+\lambda_{0}\frac{x}{L})xdx}{\int (\lambda_{o}+\lambda_O\frac{x}{L})dx}[/tex]
Solving with limits form (0,L) we get
[tex]COM_{rod}=\frac{\int_{0}^{L}(\lambda_{o}xdx+\frac{\lambda_{o}x^{2}dx}{L})}{\int_{0}^{L}(\lambda_{o}dx+\frac{\lambda_{o}xdx}{L})}\\\\COM_{rod}=\frac{\frac{\lambda L^{2}}{2}+\frac{\lambda L^{3}}{3L}}{\lambda L+\frac{\lambda L^{2}}{2L}}\\\\COM_{rod}=\frac{\frac{L}{2}+\frac{L}{3}}{1+\frac{1}{2}}\\\\COM_{rod}=\frac{5L}{9}[/tex]
